Complete question:
At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10⁵ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.
[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]
Answer:
The time it will take the particle to pass through point (P) again is 1.639 ns.
Explanation:
F = qvB
Also;

solving this two equations together;

where;
m is the mass of electron = 9.11 x 10⁻³¹ kg
q is the charge of electron = 1.602 x 10⁻¹⁹ C
B is the strength of the magnetic field = 3.47 x 10⁻³ T
substitute these values and solve for t

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.
They would be an expert in astronomy
astronomy:the branch of science that deals with celestial objects
Answer:
2.145×10^-10 V or 0.2145nV
Explanation:
From hf=eV
h= Plank's constant = 6.6×10^-34JS
f= frequency of the electromagnetic wave = 5.2×10^4 Hz
e= electronic charge= 1.6×10^-19 C
V= voltage
V= hf/e
V= 6.6×10^-34JS × 5.2×10^4 Hz/ 1.6×10^-19 C
V= 2.145×10^-10 V or 0.2145nV
Therefore the voltage created is 2.145×10^-10 V or 0.2145nV
Answer:
I think its structural plasticity.